/* permute.c */
/*    CAUTION: N factorial permutations is a bunch for N > 9
 *    	       The precise definition of a bunch is 3,628,800 up
 *             7! =   5,040   8! =  40,320    9! = 362,880
 *
 *    INPUT:   vector - input and output vector that is permuted
 *                      This must be set up befor the first call
 *                      and then used as the next permutation on
 *                      each return
 *
 *    INIT:    permuters[0] != 0    typ = n+1
 *
 *    OUTPUT:  vector - the next permuted vector
 *               if_last   = FALSE  if more permutations to be returned
 *                         = TRUE   if the last permutation is being
 *                                  returned (done)
 */
#define TRUE 1
#define FALSE 0

int permute(int n, int vector[], int permuters[]) {
  int i,j,k,t;
  int if_last = FALSE;

  if(n <= 1)return if_last=TRUE;
  if(permuters[0] != 0){
	 /* Initialization ( user must set up  vector  before first call */
    for(i=0;i<n;i++) permuters[i]=i;
  }
  else {
  /* find last in sequence of permuters 0's,reset permuters(k) to k as we go*/

    k = 0;
    while( permuters[k+1] == 0 && k<n){
      k++;
      permuters[k] = k;
    }

    /* reverse sequence  in  0 .. k */
    for(i=0;i<=k/2;i++){
      j = k - i;
      t = vector[i];
      vector[i] = vector[j];
      vector[j] = t;
    }

         /*  swap next higher item and decrement pointer */
    k++;
    permuters[k]--;
    j = permuters[k];
    t = vector[k];
    vector[k] = vector[j];
    vector[j] = t;

     /*  find out if this is the last permutation */
    k = n-1;
    if(permuters[k] == 0){
      while(k>0){
        k--;
        if(permuters[k] != 0){
          return if_last;
        }
      }
      if_last = TRUE;
    }
  }
  return if_last;
} /* end permute.c */

int permuted(int n, double vector[], int permuters[]) {
  int i,j,k;
  double t;
  int if_last = FALSE;

  if(n <= 1)return if_last=TRUE;
  if(permuters[0] != 0){
    /* Initialization ( user must set up  vector  before first call */
    for(i=0;i<n;i++) permuters[i]=i;
  }
  else {
  /* find last in sequence of permuters 0's,reset permuters(k) to k as we go*/

    k = 0;
    while( permuters[k+1] == 0 && k<n){
      k++;
      permuters[k] = k;
    }

    /* reverse sequence  in  0 .. k */
    for(i=0;i<=k/2;i++){
      j = k - i;
      t = vector[i];
      vector[i] = vector[j];
      vector[j] = t;
    }

         /*  swap next higher item and decrement pointer */
    k++;
    permuters[k]--;
    j = permuters[k];
    t = vector[k];
    vector[k] = vector[j];
    vector[j] = t;

     /*  find out if this is the last permutation */
    k = n-1;
    if(permuters[k] == 0){
      while(k>0){
        k--;
        if(permuters[k] != 0){
          return if_last;
        }
      }
      if_last = TRUE;
    }
  }
  return if_last;
} /* end permuted */
/* end permute.c */